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ira [324]
3 years ago
15

A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The

concentration of CaCl2 in this solution is ________ molal. A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl2 in this solution is ________ molal. 5.70 0.214 0.569 63.2 1.76
Chemistry
1 answer:
polet [3.4K]3 years ago
3 0

Answer:

The concentration of CaCl2 in this solution is 0.563 M

Explanation:

<u>Step 1:</u> Data given

mass of CaCl2 = 23.7 grams

Molar mass of CaCl2 = 110.98 g/mol

mass of water = 375 grams

Molar mass of water = 18.02 grams

Volume of the solution = V

<u>Step 2:</u> Calculate the mass of solution

Mass of the solution = Mass of solute + Mass of solvent

Mass of the solution = 23.7 grams + 375 grams

Mass of the solution = 398.7 grams

<u>Step 3:</u> Calculate the volume

Volume = mass / density

Volume = 398.7 grams / 1.05 mg/mL

Volume = 379.71 mL = 0.3791 L

<u>Step 4:</u> Calculate moles of CaCl2

Number of moles CaCl2 = mass CaCl2 / Molar mass CaCl2

Number of moles CaCl2 = 23.7 grams / 110.98 g/mol

Number of moles CaCl2 = 0.21355 moles

<u>Step 5:</u> Calculate the concentration of CaCl2

Concentration = Number of moles / Volume

Concentration = 0.21355 moles / 0.3791 L

Concentration = 0.563 M

The concentration of CaCl2 in this solution is 0.563 M

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BlackZzzverrR [31]

Answer:

Explanation:

How many mols do you have?

1 mol = 6.02 * 10^23 atoms

x mol = 6.25 * 10 ^32 atoms

1/x = 6.02*10^23 / 6.25 * 10^32        Cross multiply

6.02 * 10^23 * x = 1 * 6.25 * 10^32    Divide by 6.02 * 10^23

x = 6.25 * 10*32/ 6.02 ^10^23

x = 1.038 * 10^9 mols which is quite large.

Find the number of grams. (Use the value for copper on your periodic table. I will just use an approximate number.)\

1 mol of copper = 63 grams.

1.038 * 10^9 mols of copper = x

1/1.038 * 10^9 = 63/x         Cross multiply

x = 1.038 * 10^9 * 63

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Answer:

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You measure 2.34 g of K2Cr2O7 into a volumetric flask. You dilute the K2Cr2O7 with water to a volume of 250 mL. It takes 35.7 mL
natta225 [31]

Answer:

0.191M of Na₂C₂O₄ is the concentration of the original Na₂C₂O₄ solution

Explanation:

The reaction of potassium dichromate, K₂Cr₂O₇ with sodium oxalate, Na₂C₂O₄ in the presence of acid H⁺ is:

K₂Cr₂O₇ + 3Na₂C₂O₄ + 14H⁺ → 2Cr³⁺ + 6CO₂ + 7H₂O + 6Na⁺ + 2K⁺

<em>Thus, 1 mole of K₂Cr₂O₇ reacts with 3 moles of Na₂C₂O₄</em>

Moles of 2.34g of K₂Cr₂O₇ (Molar mass: 294.185g/mol):

2.34g K₂Cr₂O₇ ₓ (1mol / 294.185g) = 7.954x10⁻³ moles K₂Cr₂O₇

In 250mL = 0.250L:

7.954x10⁻³ moles K₂Cr₂O₇ / 0.250L = 0.0318M K₂Cr₂O₇

Moles in 35.7mL = 0.0357L of this solution are:

0.0357L ₓ (0.0318mol / L) = <em>1.136x10⁻³ moles K₂Cr₂O</em>₇ in solution. As 1 mole of K₂Cr₂O₇ reacts with 3 moles of Na₂C₂O₄, to titrate the moles of K₂Cr₂O₇ in solution you need:

1.136x10⁻³ moles K₂Cr₂O₇ × (3 moles Na₂C₂O₄ / 1 mole K₂Cr₂O₇) =

<em>3.408x10⁻³ moles of Na₂C₂O₄</em>

In 17.8mL = 0.0178L:

3.408x10⁻³ moles of Na₂C₂O₄ / 0.0178L =

<h3>0.191M of Na₂C₂O₄ is the concentration of the original Na₂C₂O₄ solution</h3>

<em />

3 0
3 years ago
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